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Exploring Mixtures and their Separation - Interpret solubility graphs and relate mixture behaviour to everyday contexts

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Solubility is the maximum amount of solute that can be dissolved in 100 g100 \text{ g} of a solvent at a specific temperature. It generally increases with temperature for solids in liquids, which can be visualized using a solubility curve where the x-axis represents temperature and the y-axis represents solubility.

A graph showing a curve representing solubility increasing as temperature increases.
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A saturated solution contains the maximum amount of solute possible at a given temperature. If the temperature is lowered, the solubility decreases, and the excess solute precipitates out as crystals.

Diagram
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The solubility of different salts varies at the same temperature. For example, at 20∘C20^{\circ}\text{C}, Sodium Chloride (NaClNaCl) has a different solubility compared to Potassium Chlorate (KClO3KClO_3). This behavior allows for the separation of mixtures through fractional crystallization.

Graph comparing the flat solubility curve of NaCl with the steep curve of KClO3.
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Everyday applications of solubility include the preparation of syrup (saturated sugar solutions) and the production of salt from seawater through evaporation, where the concentration of salt eventually exceeds its solubility limit.

Diagram

📐Formulae

Mass by mass percentage of a solution=(Mass of soluteMass of solution)×100\text{Mass by mass percentage of a solution} = \left( \frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100

Mass of solution=Mass of solute+Mass of solvent\text{Mass of solution} = \text{Mass of solute} + \text{Mass of solvent}

Mass by volume percentage of a solution=(Mass of soluteVolume of solution)×100\text{Mass by volume percentage of a solution} = \left( \frac{\text{Mass of solute}}{\text{Volume of solution}} \right) \times 100

Mass of crystals formed=Solubility at higher temperature−Solubility at lower temperature\text{Mass of crystals formed} = \text{Solubility at higher temperature} - \text{Solubility at lower temperature}

💡Examples

Problem 1:

A solution contains 50 g50 \text{ g} of sugar dissolved in 450 g450 \text{ g} of water. Calculate the concentration in terms of mass by mass percentage of the solution.

Solution:

Mass of solute (sugar) = 50 g50 \text{ g}. Mass of solvent (water) = 450 g450 \text{ g}. Total mass of solution = 50+450=500 g50 + 450 = 500 \text{ g}. Using the formula: Concentration=(50500)×100=10%\text{Concentration} = \left( \frac{50}{500} \right) \times 100 = 10\%

Explanation:

To find the mass percentage, we first determine the total mass of the solution by adding the solute and solvent. We then divide the mass of the solute by the total mass and multiply by 100100.

Problem 2:

The solubility of Potassium Nitrate is 62 g62 \text{ g} at 40∘C40^{\circ}\text{C} and 32 g32 \text{ g} at 20∘C20^{\circ}\text{C}. If a saturated solution containing 100 g100 \text{ g} of water is cooled from 40∘C40^{\circ}\text{C} to 20∘C20^{\circ}\text{C}, what mass of crystals will be deposited?

Solution:

Mass of solute in saturated solution at 40∘C=62 g40^{\circ}\text{C} = 62 \text{ g}. Mass of solute in saturated solution at 20∘C=32 g20^{\circ}\text{C} = 32 \text{ g}. Mass of crystals deposited: 62−3230\begin{array}{r} 62 \\ - 32 \\ \hline 30 \end{array} The mass of crystals deposited is 30 g30 \text{ g}.

Explanation:

When a saturated solution is cooled, its capacity to hold solute decreases. The difference between the solubility at the higher temperature and the lower temperature represents the amount of solute that can no longer remain dissolved and precipitates out as crystals.

Problem 3:

A student dissolves 40 g40 \text{ g} of a salt in 100 g100 \text{ g} of water at 60∘C60^{\circ}\text{C}. According to the provided graph, the solubility at 60∘C60^{\circ}\text{C} is 60 g60 \text{ g} and at 20∘C20^{\circ}\text{C} is 25 g25 \text{ g}. If the solution is cooled to 20∘C20^{\circ}\text{C}, how many grams of salt will crystallize out?

Graph showing a point moving from (60, 40) down to the solubility curve at (20, 25).

Solution:

Initial dissolved salt=40 g\text{Initial dissolved salt} = 40 \text{ g} Solubility at 20∘C=25 g\text{Solubility at } 20^{\circ}\text{C} = 25 \text{ g} Mass of crystals formed=40 g−25 g=15 g\text{Mass of crystals formed} = 40 \text{ g} - 25 \text{ g} = 15 \text{ g} Therefore, 15 g15 \text{ g} of salt will crystallize.

Explanation:

Since the student only dissolved 40 g40 \text{ g} initially (which is less than the 60 g60 \text{ g} max at 60∘C60^{\circ}\text{C}), the solution was unsaturated. Upon cooling to 20∘C20^{\circ}\text{C}, the water can only hold 25 g25 \text{ g}. The remaining 15 g15 \text{ g} must leave the liquid state.

Problem 4:

Based on the solubility graph for Salt X, what is the minimum temperature required to dissolve 50 g50 \text{ g} of the salt in 100 g100 \text{ g} of water? The solubility follows the linear equation S=0.5×T+20S = 0.5 \times T + 20.

A linear solubility graph showing the intersection at 50g and 60 degrees Celsius.

Solution:

We are given S=50 gS = 50 \text{ g}. Using the equation: 50=0.5×T+2050 = 0.5 \times T + 20 50−20=0.5×T50 - 20 = 0.5 \times T 30=0.5×T30 = 0.5 \times T T=300.5=60T = \frac{30}{0.5} = 60 The minimum temperature required is 60∘C60^{\circ}\text{C}.

Explanation:

To dissolve a specific mass, the temperature must be high enough such that the solubility limit is equal to or greater than that mass. By solving the solubility-temperature relation, we find the exact threshold.